24点问题

计算24点很简单的小程序，这一段用于解决不改变顺序的24点计算。不过最后总是有一些冗余的括号，这里为了通过SB的POJ3983，只写了一个去除最外层括号的函数，因为对这类问题的完善不太热心，就不关心去括号的事了

编译环境：G++4.6 under Fedora

/**
This piece is used to solve the POJ3983.It's another 24-point game.
The code piece is oriented from the Beauty of Programming, yet with adjustment
to solve 3983 and to make it more like C++ style.Optimization still needed.

Bug Log: Declared LIMEN as int, and equal() never returned true.
**/
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <string>
#include <assert.h>
using namespace std;

//Judge whether two double value is equal
bool equal(double,double);

//Try an operator and deal with num array and expressions
inline bool tryExp(double *num,string *exp,
        const double &a,const double &b,
        const string &expA,const string &expB,
        int i,int n,char opt);
//Recursive body of the solution,return true and stop if solution found
bool findExp(double *,string*,int);

//Test functions
void 
printExp(string *exp,int n) {
    for(int i=0;i<n;i++)
        cout<<exp[i]<<endl;
}

//Judge whether two double value is equal
bool 
equal(double a,double b) {
    const double LIMEN=0.0001;
    return fabs(a-b)<LIMEN;
}

inline bool 
tryExp(double *num,string *exp,
    const double &a,const double &b,
    const string &expA,const string &expB,
    int i,int n,char opt) {
    switch(opt) {
        case '+':num[i]=a+b;break;
        case '-': num[i]=a-b;break;
        case '*':num[i]=1.0*a*b;break;
        case '/':num[i]=1.0*a/b;break;
    }
    exp[i]='('+expA+opt+expB+')';
    return findExp(num,exp,n-1);
}
bool 
findExp(double *num,string *exp, int n) {
    assert(n<=4);
    assert(n>0);
    if(n==1) {
        if(equal(num[0],24))
            return true;
        else
            return false;
    }

    for(int i=0;i<n-1;i++) {        
        double a=num[i];
        double b=num[i+1];
        string expA=exp[i];
        string expB=exp[i+1];
        //Adjust the remain elements
        for(int j=i+1;j<n-1;j++) {
            exp[j]=exp[j+1];
            num[j]=num[j+1];
        }
        if(tryExp(num,exp,a,b,expA,expB,i,n,'+'))
            return true;
        if(tryExp(num,exp,a,b,expA,expB,i,n,'-'))
            return true;
        if(tryExp(num,exp,a,b,expA,expB,i,n,'*'))
            return true;
        if(!equal(b,0) && tryExp(num,exp,a,b,expA,expB,i,n,'/'))
            return true;
        //Adjust the array again and make process above never happened^_^
        for(int j=n-1;j>i;j--) {
            num[j]=num[j-1];
            exp[j]=exp[j-1];
        }
        num[i]=a;
        num[i+1]=b;
        exp[i]=expA;
        exp[i+1]=expB;
    } 
    return false;
}

#include <cstring>
void removeOutsideBrackets(string &exp){
    char s[50];
    strcpy(s,exp.c_str());
    int len=strlen(s);
    if(s[0]=='(' && s[len-1]==')')
        strncpy(s,s+1,len-1);
    s[len-2]=0;
    exp=s;
}

int main()
{
    double num[4];
    string exp[4];
    //Input the initial numbers. As we need expression form,
 //here input string and transform that to number later 
 //will hit two birds with one stone.
    for(int i=0;i<4;i++) {
        cin>>exp[i];
        num[i]=atoi(exp[i].c_str());
    }
    if(findExp(num,exp,4)) {
        removeOutsideBrackets(exp[0]);
        cout<<exp[0]<<endl;
    }
    return 0;
}